Characterizing Finite Quasisimple Groups by Their Complex Group Algebras

CHARACTERIZING FINITE QUASISIMPLE GROUPS BY THEIR COMPLEX GROUP ALGEBRAS

HUNG NGOC NGUYEN AND HUNG P. TONG-VIET

Abstract. A ﬁnite group L is said to be quasisimple if L is perfect and L/Z(L) is nonabelian simple, in which case we also say that L is a cover of L/Z(L). It has been proved recently [12] that a quasisimple classical group L is uniquely determined up to isomorphism by the structure of CL, the complex group algebra of L, when L/Z(L) is not isomorphic to PSL3(4) or PSU4(3). In this paper, we establish the similar result for these two open cases and also for covers with nontrivial center of simple groups of exceptional Lie type and sporadic groups. Together with the main results of [14, 16], we obtain that every quasisimple group except covers of the alternating groups is uniquely determined up to isomorphism by the structure of its complex group algebras.

1. Introduction For a finite group G, let CG denote the complex group algebra of G over the field of complex numbers; i.e. CG is the set of formal sums X { agg | ag ∈ C} g∈G equipped with natural rules for addition, multiplication, and scalar multiplication. As it is well known that CG is isomorphic to the direct sum of matrix algebras over C whose dimensions are exactly the degrees of irreducible complex representations of G, the study of complex group algebras and the relation to their base groups plays an important role in group representation theory. A fundamental question in representation theory of finite groups is whether one can recover a group or some of its properties from its complex group algebra. For instance, it was proved by M. Isaacs [6] that if two groups have isomorphic complex group algebras and one of them is nilpotent, then the other one is also nilpotent. The corresponding problem for solvability seems very difficult and is still open (see [10, Problem 11.8]).

Date: May 2, 2012. 2010 Mathematics Subject Classiﬁcation. Primary 20C33, 20C15. Key words and phrases. Complex group algebras, quasisimple groups, exceptional groups, groups of Lie type, sporadic groups. 1 2 H.N. NGUYEN AND H.P. TONG-VIET

In general, a solvable group is not determined uniquely up to isomorphism by its complex group algebra. Especially for abelian groups and p-groups, the probability that two nonisomorphic groups have the same complex group algebra is fairly “high”. On the other hand, it is believed that nonabelian simple groups or more generally quasisimple groups have a stronger connection with their complex group algebras. Conjecture 1.1. Every finite quasisimple group is determined uniquely up to isomorphism by the structure of its complex group algebra. We recall that a finite group L is said to be quasisimple if L is perfect and L/Z(L) is nonabelian simple, in which case we also say that L is a perfect central cover or simply a cover of L/Z(L). In [14, 15, 16], the second author has confirmed the conjecture for all nonabelian simple groups. It has been proved recently by the first author [12] that a quasisimple classical group L is uniquely determined up to isomorphism by the structure of CL if L/Z(L) is not isomorphic to PSL3(4) or PSU4(3). In this paper, we establish the similar result for these two open cases and moreover for all covers with nontrivial center of simple groups of exceptional Lie type and sporadic groups. In particular, we prove the following in Section3.

Theorem 1.2. Let G be a ﬁnite group and let L be the Schur cover of E6(q), E7(q), 2 ∼ ∼ or E6(q), where q is a prime power. If CG = CL then G = L.

The exceptional covers of simple groups of exceptional Lie type such as 2 · F4(2), 2 2 2 3·G2(3), 2·G2(4), {2, 2 }·Sz(8), and {2, 3, 2 , 6}· E6(2) are treated in Section4. The following theorem, which confirms the conjecture for all covers of sporadic groups, will be proved in Section5. Theorem 1.3. Let G be a finite group and let L be the a cover of a sporadic simple ∼ ∼ group. If CG = CL then G = L. As a consequence of our results, we obtain: Corollary 1.4. Every finite quasisimple group except possibly covers of the alternating groups is uniquely determined up to isomorphism by the structure of its complex group algebra. Proof. This is a consequence of Theorems 1.2, 1.3, 4.1 and 6.1, and the main results of [12, 14, 16]. We now describe briefly some ideas in the proofs. Let cd(G) denote the set of all irreducible character degrees of G. If G is a finite group and L is a quasisimple group having the same complex group algebra then certainly they have the same set of character degrees. That is, cd(G) = cd(L). Moreover, as L is quasisimple, it has a unique linear character and and hence G also has a unique linear character. It follows that G is perfect and therefore if M is a maximal normal subgroup of G then G/M COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 3 is a nonabelian simple group. Let Schur(L/Z(L)) denote the Schur cover of L/Z(L). Since every cover of L/Z(L) is a quotient of Schur(L/Z(L)), we have cd(G/M) ⊆ cd(G) = cd(L) ⊆ cd(Schur(L/Z(L))). We will use this condition together with some others if necessary to force two nonabelian simple groups G/M and L/Z(L) to be isomorphic. This basically eliminates the involvement of all nonabelian simple groups other than L/Z(L) in the structure of G. ∼ Note that as CG = CL, we also have |G| = |L|. It then follows that |M| = |Z(L)| ≤ |Mult(L/Z(L))|, where Mult(L/Z(L)) is the Schur multiplier of L/Z(L). Using this together with the facts that G/M ∼= L/Z(L) and cd(G) ⊆ cd(Schur(L/Z(L))), we prove in Section2 that G is uniquely determined by L/Z(L) and the order of G whenever L is not a cover of PSL3(4) or PSU4(3). As the Schur multipliers of these two simple groups are large, we need a closer analysis of the character degrees of their covers and this will be done in Section6.

2. Preliminaries We first define some notation that will be used throughout the paper. For a finite group G, let d(G) and b(G) denote the smallest and largest, respectively, degree of nontrivial irreducible characters of G. The Schur multiplier of G is denoted by Mult(G). If G is perfect, it is well-known that G has a unique Schur cover, which we denote by Schur(G). We write Irr(G) and cd(G) for the set of all complex irreducible characters of G and the set of all irreducible character degrees of G, respectively. By π(G) we mean the set of all prime divisors of the order of G. We will prove in this section some useful facts that will be needed in the proofs of the main theorems. We first recall two important lemmas. The first one is due to A. Moretóand the second one is due to M. Bianchi, D. Chillag, M.L. Lewis, and E. Pacifici. Lemma 2.1 ([11], Lemma 4.2). Let S be a nonabelian simple group. Then there exists a non-principal irreducible character of S that extends to Aut(S). Lemma 2.2 ([1], Lemma 5). Let N = S × · · · × S, a direct product of k copies of a nonabelian simple group S, be a minimal normal subgroup of G. If χ ∈ Irr(S) extends to Aut(S), then χk extends to G. The following is a key result of [12]. Lemma 2.3 ([12], Lemma 7.4). Let S be a nonabelian simple group. Let G be a perfect group and M ¡ G such that G/M ∼= S and |M| ≤ |Mult(S)|. Then, for every nonnegative integer i, G/M (i) is isomorphic to a quotient of Schur(S). Using this lemma, we can prove an extension of [12, Lemma 7.5]. 4 H.N. NGUYEN AND H.P. TONG-VIET

Lemma 2.4. Let S be a nonabelian simple group different from an alternating group of degree greater than 13. Let G be a perfect group and M ¡ G such that G/M ∼= S, |M| ≤ |Mult(S)|, and cd(G) ⊆ cd(Schur(S)). Then G is isomorphic to a quotient of Schur(S). Proof. The proof is very similar to that of [12, Lemma 7.5]. Indeed, the lemma is already proved there for simple groups of Lie type. Therefore, we can assume that S is a sporadic group or an alternating group of degree less than 14. If M is solvable then we are done by Lemma 2.3. So it remains to consider the case when M is nonsolvable. Let M (i) denote the ith derived subgroup of M. Since M is nonsolvable, there is a nonnegative integer i such that M (i) = M (i+1) > 1. Let N ≤ M (i) be a normal subgroup of G so that M (i)/N ∼= T k, a direct product of copies of a nonabelian simple group T . By Lemma 2.1, the group T has a non-principal irreducible character ϕ that extends to Aut(T ). Lemma 2.2 then implies that ϕk extends to G/N. Therefore, using Gallagher’s lemma [5, Corollary 6.17], we have ϕkχ ∈ Irr(G/N) for every χ ∈ Irr(G/M (i)). In particular, (ϕkχ)(1) = ϕ(1)kχ(1) ∈ cd(G/N) ⊆ cd(G) ⊆ cd(Schur(S)). Take χ to be an irreducible character of the largest degree of S. By Lemma 2.3, the simple group S is a quotient of G/M (i) and hence χ can be considered as an irreducible character of G/M (i). We deduce that Schur(S) has a character degree which is a proper multiple of the largest character degree of S. However, this cannot happen by a routine check from [2]. Lemma 2.5. Let S be a nonabelian simple group different from an alternating group of degree greater than 13. Assume that S is different from S = PSL3(4) and PSU4(3). Let G be a perfect group and M ¡ G such that G/M ∼= S, |M| ≤ |Mult(S)|, and cd(G) ⊆ cd(Schur(S)). Then G is uniquely determined up to isomorphism by S and the order of G. Proof. The lemma is already proved in [12, Lemma 7.6] for simple group of Lie type. So it remains to consider the case where S is a sporadic group or an alternating group of degree less than 14. In particular, we can assume that the Schur multiplier Mult(S) of S is cyclic. As proved in Lemma 2.4 that G is isomorphic to a quotient of Schur(S), we assume G ∼= Schur(S)/Z, where Z ≤ Mult(S). Since G/M ∼= S, we then deduce that |Z| = |Mult(S)|/|M| = |Schur(S)|/|G|. Since the cyclic group Mult(S) has a unique subgroup of order |Schur(S)|/|G|, the subgroup Z is uniquely determined by S and |G|. The lemma follows. COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 5

2 3. The Schur covers of E6(q), E6(q) and E7(q) 2 Proposition 3.1. Let H be the simple group E6(q), E6(q) or E7(q), where q is a power of a prime p. Let S be a nonabelian simple group such that cd(S) ⊆ cd(Schur(H)). Then S ∼= H. This proposition will be proved in Lemmas 3.6, 3.7 and 3.8. We first recall some known results. Lemma 3.2 (Zsigmondy’s theorem [17]). Let q and n be integers with q ≥ 2 and n ≥ 3. Then qn − 1 has a prime divisor ` such that ` does not divide qm − 1 for every m < n, unless (q, n) = (2, 6). Moreover if ` | qk − 1 then n | k. n Such an ` is called a primitive prime divisor of q − 1. We denote by `n(q) the smallest primitive prime divisor of qn − 1. Recall that b(G) and d(G) are the largest and smallest, respectively, degree of nontrivial irreducible character degree of G. We need the following lower bound of b(An). n−1 Lemma 3.3 ([14], Lemma 2.2). If n ≥ 10, then b(An) ≥ 2 The following lemma is straightforward and we skip the proof. Lemma 3.4. Let S be a nonabelian simple group and G be a group. If cd(S) ⊆ cd(G) then the following hold: (i) d(S) ≥ d(G), (ii) b(S) ≤ b(G), (iii) π(S) ⊆ π(G). Suppose that q = pa where a ∈ N. We prove in the following lemma that S is a simple group of Lie type in characteristic p under the assumption of Proposition 3.1. Lemma 3.5. Assume the hypothesis of Proposition 3.1. Then S is a simple group of Lie type in the same characteristic p as that of H. Proof. Since S is simple, we will employ the classification of finite simple groups and eliminate all possibilities of S except when S is a simple group of Lie type in characteristic p. (i) Assume that S is a sporadic simple group or the Tits group. Assume first that ∼ 2 H = E6(q) or E6(q). The smallest nontrivial degrees of irreducible characters of exceptional groups of Lie type are determined by Lübeck [7]. In particular, we have d(Schur(H)) ≥ q(q4 + 1)(q6 − q3 + 1). ∼ 2 ∼ If q = 2 or 3 but H 6= E6(2), then |Mult(H)| = 1 and so Schur(H) = H is simple ∼ 2 and the result follows from [14, Proposition 3.2]. Assume now that H = E6(2). Again by [7], we see that d(Schur(H)) = 1938 and thus by Lemma 3.4(i) we have d(S) ≥ d(Schur(H)) = 1938. 6 H.N. NGUYEN AND H.P. TONG-VIET

By checking the smallest character degrees of sporadic simple groups and the Tits group in [2], we obtain 0 0 S ∈ {O N, Ly, F i24,B,M}. For each possibility of S, we can verify that π(S) * π(Schur(H)), and this contradicts Lemma 3.4(iii). Now we can assume that q ≥ 4. Then d(Schur(H)) ≥ 4145924 > d(S) for all sporadic groups and the Tits group S, a contradiction by Lemma 3.4(i). ∼ Similar arguments apply to H = E7(q). Recall that |Mult(H)| = gcd(2, q − 1). Therefore, if q = 2 then Schur(H) is simple and the result follows from [14, Proposi- tion 3.2]. Hence we assume that q ≥ 3. By [7] and [2] , we have d(Schur(H)) ≥ q(q14 − 1)(q4 − q2 + 1)/(q2 − 1) ≥ 120933749 > d(S) for all sporadic simple groups or the Tits group S. This violates Lemma 3.4(i). ∼ (ii) Next, we assume that S = An with degree n ≥ 5. As cd(S) ⊆ cd(Schur(H)), we deduce from Lemma 3.4(i) that d(S) ≥ d(Schur(H)). Using the result of Lubeck again, we obtain d(Schur(H)) ≥ 1938 and thus d(An) ≥ 1938, which implies that n ≥ 6 as d(A5) = 3. This, in turn, implies that d(An) = n − 1. Hence n − 1 ≥ 1938 and so n ≥ 1939. Now, Lemma 3.4 yields that

b(Schur(H)) ≥ b(An) and n − 1 = d(An) ≥ d(Schur(H)). n−1 By Lemma 3.3, we know that b(An) ≥ 2 . Therefore b(Schur(H)) ≥ 2d(Schur(H)). ∼ 2 Recall that H = E6(q), E6(q) or E7(q). It is shown in [7] that d(Schur(H)) ≥ q(q4 + 1)(q6 − q3 + 1). In particular, we have d(Schur(H)) ≥ q10. On the other hand, p b(Schur(H)) < p|Schur(H)| < q133 < q67. We deduce that q67 > 2q10 . However, this last inequality is impossible for q ≥ 2. (iii) Finally, we consider the case when S is a simple group of Lie type. If the characteristic of the underlying field of S is different from p, the group Schur(H) would have two different prime power degrees and this violates the classification of prime power irreducible character degrees of quasisimple groups by G. Malle and COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 7

A.E. Zalesskii [9, Theorem 1.1]. We conclude that the deﬁning characteristic of S must be p. ∼ Lemma 3.6. Proposition 3.1 is true when H = E6(q). Proof. We have seen from Lemma 3.5 that S is a simple group of Lie type in charac- ∼ a teristic p. Assume that H = E6(q) = E6(p ) and that the underlying ﬁeld of S has b cardinality q1 = p . As cd(S) ⊆ cd(Schur(H)) and StH (1) = |H|p is the only prime power degree of Schur(H) by [9, Theorem 1.1], we obtain 36a (3.1) |S|p = p .

We will eliminate all the possibilities of S except E6(q). 3 (1) The case when S is of type G2, F4 or D4 can be argued as follows. Suppose bm that |S|p = p and the largest positive integer x such that the primitive prime 0 divisor `bx(p) divides |S| is m . Formula (3.1) implies that bm = 36a.

By Lemma 3.4(iii), we obtain `bm0 (p) ∈ π(Schur(H)). Recall that 36 12 9 8 6 5 2 |Schur(E6(q))| = q (q − 1)(q − 1)(q − 1)(q − 1)(q − 1)(q − 1). Therefore, bm0 ≤ 12a. It follows that m0/m ≤ 1/3, which is a contradiction by a straightforward check of the order of S. 2 2 2 (2) The case when S is of type B2, G2 or F4 is eliminated as in (1) by rewriting the order of S is the form: 2 2 4 2 | B2(q1)| = q1(q1 − 1)(q1 − 1)/(q1 − 1), 2 3 6 3 | G2(q1)| = q1(q1 − 1)(q1 − 1)/(q1 − 1), and 2 12 12 4 3 | F4(q1)| = q1 (q1 − 1)(q1 − 1)(q1 − 1)/(q1 − 1).

2 (3) Assume that S is of type E6. Then the degree of the Steinberg character of 36b a S is p . This degree again must be the degree of the Steinberg character of E6(p ), 36a which is p . Therefore a = b. Now one can check that `18a(p) ∈ π(S) but it does not divide |Schur(H)|, violating Lemma 3.4(iii). ∼ b (4) The cases where S is of type E7 and E8 are eliminated similarly. If S = E7(p ) then (3.1) implies that 63b = 36a or equivalently 4a = 7b. Note that (p18b − 1) | |S|. Hence `18b(p) ∈ π(S) and so `18b(p) divides |Schur(H)|. Therefore, as 18b = 72a/7 > 9a, we must have 18b divides 12a by Zsigmondy’s theorem. However as 4a = 7b, we obtain that 12a = 21b and thus 18b divides 21b, which is impossible. 8 H.N. NGUYEN AND H.P. TONG-VIET ∼ For the case S = E8(q), we have 120b = 36a and thus 10b = 3a. We ﬁrst have that `24b(p) ∈ π(S) and thus by Zsigmondy’s theorem, we deduce that 24b divides 8a, 9a or 12a since 24b > 20b = 6a. Assume that 24b divides 8a. Then 120b = 36a divides 40a and this is impossible. The remaining cases can be argued similarly.

Table 1. The smallest degree of unipotent characters of ﬁnite classical groups (see [13]).

Groups Degrees n n−1 q1 − q1 PSLn(q1), = ±1 q1 − n n (q1 − 1)(q1 − q1) Ω2n+1(q1) or PSp2n(q1) 2(q1 + 1) n n−1 (q1 − )(q1 + q1) P Ω2n(q1), = ±1 2 q1 − 1

∼ b (5) Assume that S = PSLn(q1) where q1 = p . As cd(S) ⊆ cd(Schur(E6(q))), Table1 yields n n−1 q1 − q1 χ(1) = ∈ cd(Schur(E6(q))). q1 − It follows from (3.1) that n(n − 1)b/2 = 36a and so (bn)2 ≥ n(n − 1)b = 72a ≥ 82. Thus bn ≥ 8 and therefore `bn(p) ∈ π(S) exists by Zsigmondy’s theorem. Now Lemma 3.4(iii) implies that `bn(p) ∈ π(Schur(H)). Zsigmondy’s theorem then implies that bn ≤ 12a. Therefore

n nb 12a 12 χ(1) < q1 = p ≤ p = q .

By [7, §5.8], the second smallest nontrivial degree of Schur(E6(q)) is q2(q2 + 1)(q4 + 1)(q5 − 1)(q4 − q2 + 1)/(q − 1) which is greater than q15. Therefore, χ(1) must be the smallest nontrivial character degree of Schur(E6(q)). That is qn − n−1q 1 1 = q(q4 + 1)(q6 + q3 + 1). q1 −

Comparing the p-parts of both sides, we obtain q1 = q. It follows from (3.1) that n(n−1)/2 36 q = |S|p = q , which implies that n = 9. But then `7(q) ∈ π(S)−π(Schur(H)), contradicting Lemma 3.4(iii). ∼ ± bn(n−1) (6) Assume S = P Ω2n(q1) with n ≥ 4. We have |S|p = p . We deduce from (3.1) that bn(n − 1) = 36a and thus (2b(n − 1))2 > bn(n − 1) = 36a ≥ 62, COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 9 therefore 2b(n − 1) > 6 and hence `2b(n−1)(p) ∈ π(S) exits. So, by Zsigmondy’s theorem, we have 2b(n − 1) ≤ 12a. By Table1, S has an irreducible character χ with n n−1 (q1 − )(q1 + q1) 2(n−1) 2b(n−1) 12a 12 χ(1) = 2 < q1 = p ≤ p = q . q1 − 1

Using the same argument as in item (5), we obtain that q1 = q. Now (3.1) implies that n(n − 1) = 36, which is impossible. ∼ (7) Assume that S = Ω2n+1(q1) or PSp2n(q1) with n ≥ 2. Arguing as in (5), we get bn2 = 36a and (qn − 1)(qn − q ) 1 1 1 = q(q4 + 1)(q6 + q3 + 1). 2(q1 + 1) If p is odd, then it follows from latter equation that q1 = q and thus b = a, which implies that n = 6. But then the equation above has no solutions. If p = 2, then 2 q = q1/2 which implies that a = b − 1. It follows from the equation bn = 36a that (a + 1)n2 = 36a. As 36a = (a + 1)n2 > an2, we obtain 2 ≤ n ≤ 5. Furthermore, as (a + 1)n2 = 36a, we deduce that a(36 − n2) = n2, which implies that 36 − n2 ≤ n2 and so n2 ≥ 18. Thus n ≥ 5. Therefore we conclude that n = 5. However the equality 2 (a + 1)n = 36a cannot happen. ∼ 2 Lemma 3.7. Proposition 3.1 is true when H = E6(q). Proof. Arguing as in Lemma 3.6, we obtain 36a (3.2) |S|p = p . Recall that if q > 2, then 2 36 12 9 8 6 5 2 |Schur( E6(q))| = q (q − 1)(q + 1)(q − 1)(q − 1)(q + 1)(q − 1) and if q = 2, then 2 38 12 9 8 6 5 2 |Schur( E6(2))| = 2 (2 − 1)(2 + 1)(2 − 1)(2 − 1)(2 + 1)(2 − 1). 2 3 2 2 (1) The case when S is of type G2, B2, D4, G2 or F4 can be argued as in items (1) and (2) of the proof of Lemma 3.6. So we skip the details.

(2) The case when S is of type F4,E6,E7 or E8 can be argued as in items (3) and (4) of the proof of Lemma 3.6 with the observation that q9 − 1 does not divide 2 |Schur( E6(q))|. Again we skip the details. ∼ b (3) Assume S = PSLn(p ), n ≥ 2. By inspecting the list of character degrees of Schur(H) given in [8], we deduce that if χ ∈ Irr(Schur(H)) with χ(1) 6= StH (1), 25a then χ(1)p, the p-part of χ(1), is at most p . By Table2, S has a degree diﬀerent b(n−1)(n−2)/2 from the degree of the Steinberg character StS with the p-part p . Hence b(n−1)(n−2)/2 ≤ 25a. Furthermore by (3.2) we have pbn(n−1)/2 = p36a, which implies 10 H.N. NGUYEN AND H.P. TONG-VIET

Table 2. Some unipotent characters of simple classical groups of Lie type

S = S(pb) Symbol p-part of degree b n−2 b(n−1)(n−2)/2 PSLn(p ) (1 , 2) p b b 0 1 2···n−2 n−1 n b(n−1)2 PSp2n(p ), Ω2n+1(p ), p > 2 1 2··· n−2 p b b(n−1)2−1 PSp2n(p ), p = 2 2 + b 0 1 2··· n−3 n−1 b(n2−3n+3) P Ω2n(p ) 1 2··· n−2 n−1 p − b 0 1 2 ··· n−2 n b(n2−3n+2) P Ω2n(p ) 1 2··· n−2 p that bn(n − 1)/2 = 36a. Multiplying both sides of the latter equation by n − 2, we obtain 36a(n − 2) = bn(n − 1)(n − 2)/2 ≤ 25an. Simplifying this inequality, we see that n ≤ 6. Now arguing as in the proof of (5) of Lemma 3.6, we deduce that `bn(p) ∈ π(S) exists and so bn ≤ 18a. Multiplying both sides of this inequality by 2, we obtain 36a = bn(n − 1)/2 ≥ 2bn, which implies that n ≥ 5. Therefore we conclude that n = 5 or 6. If n = 5, then 5b = 18a by (3.2). Hence 4b = 72a/5 > 12a. So, by Zsigmondy’s theorem we must have that `4b(p) ∈ π(S) exists and thus 4b must divide 18a = 5b, a contradiction. The case n = 6 can be argued similarly. ∼ b b 2 (4) Assume S = PSp2n(p ) or Ω2n+1(p ) with n ≥ 2. By (3.2), we have bn = 36a. If p = 2 and 2bn = 6, then bn = 3 so that n = 3, b = 1 and hence 9 = 36a, which is not possible. Thus `2bn(p) ∈ π(S) exists and so `2bn(p) ∈ π(Schur(H)). We deduce that 2bn ≤ 18a or equivalently bn ≤ 9a. Multiplying both sides by n, we obtain that bn2 = 36a ≤ 9an. Hence n ≥ 4. By Table2, S has a unipotent character ψ with b(n−1)2−b ψ(1)p ≥ p and so as in the proof of (3), we obtain bn(n − 2) ≤ 25a. Multiplying both sides by n, we have bn2(n − 2) = 36a(n − 2) ≤ 25an. Therefore 4 ≤ n ≤ 6. If n = 4 then 4b = 9a. We have 6b = 27a/2 > 13a and so `6b(p) ∈ π(S) ⊆ π(Schur(H)) exists and hence 6b divides 18a = 8b, which is impossible. Similarly, if n = 5 then 25b = 36a. We have 10b = 360a/25 = 72a/5 > 12a and so `10b(p) ∈ π(Schur(H)) exists. Hence 10b divides 18a, and so 20b must divide 36a = 25b, a contradiction. If n = 6 then b = a. However we see that `5(q) ∈ π(S) exists but `5(q) 6∈ π(Schur(H)), which contradicts Lemma 3.4(iii). ∼ b (5) Assume S = P Ω2n(p ), n ≥ 4. It follows from (3.2) that bn(n − 1) = 36a. By b(n−1)(n−2) Table2, S has a unipotent character ψ with ψ(1)p ≥ p . With the same reasoning as in item (3), we obtain b(n − 1)(n − 2) ≤ 25a. COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 11

Multiplying both sides by n and simplifying, we obtain 11n ≤ 72 and so 4 ≤ n ≤ 6. ∼ 3 ∼ 2 If n = 4 then b = 3a, which implies that S = O8(q ) and H = E6(q). We have `9(q) ∈ π(S) but `9(q) 6∈ π(Schur(H)), contradicting Lemma 3.4(iii). If n = 5 then 5b = 9a. As 8b = 72a/5 > 12a, `8b(p) ∈ π(S) ⊆ π(Schur(H)) exists and hence 8b divides 18a = 10b, which is impossible. Finally, assume that n = 6. Then 5b = 6a. As above, we have `8b(p) ∈ π(Schur(H)) and since 8b = 48a/5 > 9a, we deduce that 8b divides 10a, 12a = 10b or 18a = 15b. Obviously the last two cases cannot happen. If 8b divides 10a, then 24b divides 30a = 25b, which is again impossible.

Lemma 3.8. Proposition 3.1 is true when S = E7(q). Proof. Again, we know from Lemma 3.5 that S is a simple group of Lie type in ∼ ∼ characteristic p and we aim to show that S = E7(q). Assume that H = E7(q) = a b E7(p ) and that the underlying ﬁeld of S has cardinality q1 = p . As cd(S) ⊆ cd(Schur(H)) and StH (1) = |H|p is the only prime power degree of Schur(H) by [9, Theorem 1.1], we obtain 63a (3.3) |S|p = p .

2 3 2 2 2 (1) Assume that S is of type G2, F4, E6, D4, G2, F4 or E6. Suppose that bm 0 |S|p = p and the largest integer x such that `bx(p) divides |S| is m . Formula (3.3) implies that bm = 63a.

By Lemma 3.2, `bm0 (p) divides |Schur(E7(q))|. Recall that 63 18 14 12 10 8 6 2 |Schur(E7(q))| = q (q − 1)(q − 1)(q − 1)(q − 1)(q − 1)(q − 1)(q − 1). Therefore, Zsigmondy’s theorem implies bm0 ≤ 18a. It follows that m0/m ≤ 18/63 = 2/7, which is a contradiction by a routine check. ∼ b (2) Assume that S = E8(p ). Then (3.3) implies that 120b = 63a or equivalently 40b = 21a. Now the prime `30b(p) divides |S| but does not divide |Schur(H)|, a contradiction. (3) Arguments for the cases when S is a simple classical group are similar. We ∼ present here only the case S = PSp2n(q1) with n ≥ 2. As cd(S) ⊆ cd(Schur(E7(q))), Table1 yields n n (q1 − 1)(q1 − q1) χ(1) = ∈ cd(Schur(E7(q))). 2(q1 + 1) As in the proof of (5) of Lemma 3.6, we deduce that 2bn ≤ 18a. Therefore 2n 2bn 18a 18 χ(1) ≤ q1 = p ≤ p = q . 12 H.N. NGUYEN AND H.P. TONG-VIET

18 By [7, §5.9], the second smallest nontrivial degree of Schur(E7(q)) is greater than q . Hence χ(1) must be the smallest nontrivial degree of Schur(E7(q)). That is n n 14 4 2 (q1 − 1)(q1 − q1) q(q − 1)(q − q + 1) = 2 . 2(q1 + 1) q − 1

Comparing the p-parts of both sides, we see that either q1 = q or q1/2 = q. Assume n2 63 ﬁrst that q1 = q. It follows from (3.3) that q = q , which is impossible. Assume now that q1/2 = q or equivalently b − 1 = a. Formula (3.3) implies that bn2 = (a + 1)n2 = 63a. Firstly, we have 63a = (a + 1)n2 > an2, which implies that n ≤ 7. Secondly, as (a + 1)n2 = 63a, we deduce that a(63 − n2) = n2. Hence 63 − n2 ≤ n2, which implies that 2n2 ≥ 63 and thus n ≥ 5. Therefore 5 ≤ n ≤ 7. For each possibility of n, we see 2 2 that the equality a(63 − n ) = n cannot occur. We are now ready to prove the main result of this section. Proof of Theorem 1.2. Recall that G is a ﬁnite group and L is the Schur cover of 2 ∼ E6(q), E7(q) or E6(q) such that CG = CL. In particular, G is perfect, |G| = |L|, and cd(G) = cd(L). Let H := L/Z(L). It follows by Proposition 3.1 that if M is a maximal normal subgroup of G, then G/M ∼= H. As |G| = |L| = |Schur(H)|, it follows that |M| = |Mult(H)|. Now we observe that G ∼ and L both satisfy the hypothesis of Lemma 2.5 and so G = L.

4. Exceptional covers For the covers of simple groups of exceptional Lie type, besides all simple groups 2 considered in [14] and the Schur covers of E6(q), E7(q) or E6(q) considered in Sec- tion3, we are left with a few exceptional covers. The aim of this section is to prove the following. Theorem 4.1. Let G be a ﬁnite group and let 2 2 2 L ∈ {2 · F4(2), 3 · G2(3), 2 · G2(4), {2, 2 }· Sz(8), {2, 3, 2 , 6}· E6(2)}. ∼ ∼ If CG = CL then G = L. Let C be the set of quasisimple groups mentioned in Theorem 4.1. We ﬁrst prove an analogue of Proposition 3.1 for the groups in C. Proposition 4.2. Let L ∈ C and let G be a perfect group such that |G| = |L| and cd(G) = cd(L). If M is a maximal normal subgroup of G, then G/M ∼= L/Z(L). COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 13

2 Proof. If L ∈ C is a cover of E6(2), then the result follows from Proposition 3.1. Other groups can be argued as follows. Since |G| = |L| and cd(G) = cd(L), it follows that |G/M| divides |L| and cd(G/M) ⊆ cd(L). With L ∈ {2 · F4(2), 3 · 2 G2(3), 2 · G2(4), 2 · Sz(8), 2 · Sz(8)}, it is routine to check from [2] that there is no nonabelian simple group S diﬀerent from L/Z(L) such that |S| divides |L| and cd(S) ⊆ cd(L). ∼ Proof of Theorem 4.1. The proof is similar to that of Theorem 1.2. Since CG = CL, we have cd(G) = cd(L) and G is perfect. It follows by Proposition 4.2 that G/M ∼= L/Z(L). As |G| = |L|, we deduce that |M| = |Z(L)| and hence |M| ≤ |Mult(L/Z(L))|. Now we ∼ see that G and L both satisfy the hypothesis of Lemma 2.5 and therefore G = L.

5. Covers of sporadic simple groups Let H be a sporadic simple group. By [2], we know that Mult(H) is a cyclic group which is trivial unless H appears in Table3. In this table, ˜p(G) is the smallest prime which does not divides |G|.

Table 3. Sporadic groups with nontrivial Schur multiplier

S |Mult(S)| p˜(Schur(S)) M12 2 7 M22 12 13 J2 2 11 HS 2 13 J3 3 7 McL 3 13 Ru 2 11 Suz 6 17 O0N 3 13 F i22 6 17 Co1 2 17 0 F i24 3 19 B 2 29

Proposition 5.1. Let H be a sporadic simple group. If S is a nonabelian simple ∼ ∼ ∼ group with cd(S) ⊆ cd(Schur(H)), then S = H or S = M11 and Schur(H) = 2 · M12. Proof. If H has trivial Schur multiplier then the result follows from [16]. So we assume that the Schur multiplier Mult(H) of H is nontrivial and thus H appears in 14 H.N. NGUYEN AND H.P. TONG-VIET

Table3. Using the classiﬁcation of ﬁnite simple groups, we consider the following cases: (1) S is a sporadic simple group or the Tits group. Using [2] and Lemma 3.4, it is ∼ ∼ ∼ routine to check that either S = H or S = M11 and Schur(H) = 2 · M12. ∼ (2) S = An, n ≥ 5. If n ≤ 17, then one can check that the conclusion of the proposition holds by using [2] and [3]. Thus we can assume that n ≥ 18. As {7, 11, 13, 17} ⊆ π(An), it follows by Lemma 3.4(iii) that {7, 11, 13, 17} ⊆ π(Schur(H)). Table3 then ∼ 0 implies that S = F i24 or B. Now by Lemma 3.4(i), we have

d(An) = n − 1 ≥ d(Schur(H)) ≥ 783.

Thus n ≥ 784 and hence 73 ∈ π(An), which implies that 73 ∈ π(Schur(H)), a contradiction. (3) S is a simple group of Lie type in characteristic p different from the Tits group. In this case, S has a prime power character degree pa, which is the degree of the Steinberg character of S. Now by [9, Theorem 1.1], we are in one of the following situations: ∼ a 4 5 (i) Schur(H) = 2 · M12 and p ∈ {11, 2 , 2 }; ∼ a 6 (ii) Schur(H) = 2 · J2 and p = 2 ; (iii) Schur(H) ∼= 2 · Ru and pa = 213. ∼ a b If Schur(H) = 2 · M12 or 2 · J2, then |π(Schur(H))| = 4. Using Burnside’s p q Theorem [5, Theorem 3.10] and Lemma 3.4(iii), we have 3 ≤ |π(S)| ≤ 4. Inspecting the lists of simple groups of Lie type with only three or four distinct prime divisors ∼ (see [4, Lemmas 2, 3]), we can check that cd(S) " cd(Schur(H)) unless S = H. We are left with case (3). Assume that Schur(H) ∼= 2·Ru and S is a simple group of 13 Lie type in characteristic 2 with |S|2 = 2 . We have π(Schur(H)) = {2, 3, 5, 7, 13, 29}. We consider the following subcases. ∼ f (a) S = PSLn(2 ). Then fn(n − 1)/2 = 13. Assume first that n = 2. Then ∼ 13 13 S = PSL2(2 ). However we can see that π(PSL2(2 )) * π(Schur(H)). Hence we assume that n ≥ 3. It follows that n(n − 1)/2 ≥ 3. Since 13 is prime, we deduce that f = 1 and n(n − 1)/2 = 13. But the latter equation has no integer solutions. ∼ f 2 (b) S = PSp2n(2 ), n ≥ 2. We have fn = 13 which is impossible. ∼ f (c) S = P Ω2n(2 ), n ≥ 4. Then fn(n − 1) = 13 which is impossible as n(n − 1) is always even. (d) S is one of the simple exceptional groups of Lie type defined over a field of f mf size 2 . Write |S|2 = 2 . Then mf = 13 and thus m must be odd. Checking the ∼ f orders of the simple exceptional groups of Lie type, we obtain that S = E7(2 ) and so 63f = 13, which is a contradiction. COMPLEX GROUP ALGEBRAS OF QUASISIMPLE GROUPS 15 ∼ ∼ Proof of Theorem 1.3. Assume that CG = CL, where H = L/Z(L) is a sporadic simple group. As in the proof of Theorem 1.2, we deduce that |G| = |L|, cd(G) = cd(L) and G is perfect. Let M be a maximal normal subgroup of G. Then G/M is a nonabelian simple group and cd(G/M) ⊆ cd(L) ⊆ cd(Schur(H)). By Propostion 5.1, ∼ ∼ ∼ we have either G/M = H or G/M = M11 and L = 2 · M12. If the former case occurs, then the theorem follows from Lemma 2.5. ∼ We will show that the latter case indeed cannot happen. Observe that as CG = CL, the multiplicity of every character degree of G and L are the same and thus for each character degree x of G/M, the multiplicity of x in G/M does not exceed that of x in ∼ L. However we see that the character of degree 10 has multiplicity 3 in G/M = M11 ∼ but multiplicity 2 in L = 2 · M12, a contradiction.

6. Covers of PSL3(4) and PSU4(3) It has been proved in [12] that every quasisimple classical group except possibly covers of PSL3(4) and PSU4(3) is determined up to isomorphism by the structure of its complex group algebra. The aim of this section is to prove the following result, which addresses these two cases. Theorem 6.1. Let G be a ﬁnite group and let L be a quasisimple group where L/Z(L) ∼ ∼ is PSL3(4) or PSU4(3). If CG = CL then G = L. Proof. Let M be a maximal normal subgroup of G and set H := L/Z(L). It has been shown in [12, Proposition 1.2] that G/M ∼= H. Therefore, as |G| = |L|, we obtain |M| = |Z(L)| ≤ |Mult(H)|. Moreover, since ∼ CG = CL, we have cd(G) = cd(L) ⊆ cd(Schur(H)). Now we can use Lemma 2.4 to deduce that G ∼= Schur(H)/Z for some Z ≤ Mult(H). This means that G and L are both covers of H of the same order. We now follow the notation of the covers of PSL3(4) and PSU4(3) in [2, pages 22, 52].

(1) H = PSL3(4). Recall that Mult(H) = Z4 × Z4 × Z3. If |Z(L)|= 6 4 and 12 then we are done since there is only one cover of PSL3(4) (up to isomorphism) whose center is of order |Z(L)|. If |Z(L)| = 4 then G and L are isomorphic to either 41 · PSL3(4) or ∼ 42 · PSL3(4). However, as cd(41 · PSL3(4)) 6= cd(42 · PSL3(4)), we deduce that G = L, as claimed. A similar argument works for the case |Z(L)| = 12.

(2) H = PSU4(3). Then Mult(H) = Z3 × Z3 × Z4. If |Z(L)| 6= 3, 6, and 12 then we are done as above. If |Z(L)| = 3 then G and L are isomorphic to 31 · PSU4(3) or ∼ 32 · PSU4(3). As cd(31 · PSL3(4)) 6= cd(32 · PSL3(4)), we again conclude that G = L. The cases |Z(L)| = 6 and |Z(L)| = 12 are argued similarly. 16 H.N. NGUYEN AND H.P. TONG-VIET

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Department of Mathematics, The University of Akron, Akron, Ohio 44325, United States E-mail address: [email protected]

School of Mathematics, Statistics and Computer Science, University of KwaZulu- Natal, Pietermaritzburg 3209, South Africa E-mail address: [email protected]